NCEA Level 2 Chemistry (90310) 2011 — page 1 of 4
Assessment Schedule – 2011
Chemistry: Describe thermochemical and equilibrium principles (90310)
Evidence Statement
Q
ONE
(a)(i)
Evidence
A: It has higher concentration. It has more particles in a given
volume. It has more frequent collisions. It has a higher rate of
reaction.
(ii)
Granules have greater surface area. Thus more Zn atoms are
exposed. Thus more frequent collisions occur.
(iii)
The change causes a lower rate of reaction. It lowers kinetic
energy. It causes less frequent collisions.
Thus fewer particles have EA. Thus fewer collisions are successful.
Achievement with Merit
Achievement with Excellence
TWO of:
• ONE idea.
TWO of
• A AND TWO ideas.
TWO of
• A AND full explanation
•
Greater surface area of zinc /
more atoms exposed to
collision.
•
TWO ideas.
•
TWO ideas.
•
Lower rate with Lower Ek
AND
ONE other idea.
TWO of OR α
TWO
(a) (i)
Achievement
[CH3OH]
Kc =
[CO][H2]
(ii)
Kc small OR Kc <1: reactants > products
(b)
When temperature is decreased: reaction occurs to: reduce
decrease in T: favouring the forward, exothermic reaction.
A catalyst has no effect on yield: since it increases the rates of
forward and reverse reactions equally.
When pressure is increased: reaction occurs to: reduce increase
in pressure: favouring the forward reaction which reduces the
number of moles of gas.
α AND one other
•
Correct Kc expression.
•
Correct Kc expression.
•
Reactants > products OR
reactant favoured.
•
Reactants > products OR
reactant favoured
•
TWO correct changes.
OR
ONE correct change supported
by one idea.
•
TWO correct changes
supported by one idea each.
(α)
•
Full explanation
•
Full explanation
β and one other
OR α and two others
• Correct Kc expression.
•
Full answer.
•
Full answer for temperature
and pressure.
(β)
NCEA Level 2 Chemistry (90310) 2011 — page 2 of 4
THREE
(a)(i)
TWO of:
OR
(ii)
(b) (i)
(ii)
(as acid)
(as base)
NH4+ / NH3
HPO42– / PO43–
(iii)
[H3O+] = inverse log –pH = 10–4.62 = 2.40 10–5 mol L–1
KW
1´10-14
[OH - ] =
=
= 4.17 ´10-10 mol L-1
+
-5
[H 3O ] 2.40 ´10
(c)
HNO3 + H2O  NO3– + H3O+
strong acid OR fully dissociates
produces high(er) conc. of ions
higher conductivity
(N1)
(N2)
(N3)
(N4)
HCOOH + H2O ⇌ HCOO– + H3O+
weak acid OR partially dissociates
producing low(er) conc. ions
Hence lower conductivity
(W1)
(W2)
(W3)
(W4)
KOH  K+ + OH–
Strong base OR fully soluble
produces high(er) conc. ions
higher conductivity
(K1)
(K2)
(K3)
(K4)
2β AND α
•
correct pair .
OR
ONE equation.
•
Correct pair.
AND
BOTH equations.
(β1)
•
•
TWO values.
•
THREE values.
(α2)
 FOUR values (with units).
(β2)
•
THREE codes for one solution.
•
THREE codes for each of
two solutions.
(α3)
•
 SO42– + H3O+
 H2SO4+ OH–
pH = –log [H3O+] = –log 0.0498 = 1.30
pOH = –log [OH–] = 0.60 OR
KW
1´10-14
[H 3O+ ] =
=
= 3.98 ´10-14 mol L-1
0.251
[OH - ]
pH = –log [H3O+] = –log 3.9810–14 = 14 – 0.60 = 13.4
TWO of
OR
Equation
Or strength
Or [ion]
Or conductivity
For THREE solutions.
Correct pair
AND
BOTH equations.
(β1)
FOUR codes for each of two
solutions.
(β3)
NCEA Level 2 Chemistry (90310) 2011 — page 3 of 4
FOUR
(a)(i)
(ii)
(b)(i)
(ii)
TWO of:
Endothermic: positive: value
Exothermic: bonds form OR energy released
1652 kJ
E =
= 826 kJ mol-1 (kJ)
2.00 mol
1652 kJ
= 413 kJ mol-1
4 mol
4 mol ´185 kJ
185 kJ
n(Fe) =
=
= 0.448 mol
1652 kJ
413 kJ mol-1
m(Fe) = nM= 0.448 mol  55.9 g mol-1 = 25.0 g
THREE of:
•
ONE choice with reason
OR TWO choices
•
TWO choices with ONE
reason.
•
ONE step correct (ignore sign).
•
826 or 25.0
•
ONE idea.
•
ONE idea.
•
143 or 55.6 (ignore sign).
THREE of:
•
TWO choices with reasons.
•
826 kJ AND 25.0 g.
•
ONE idea.
•
143 kJ g–1
55.6 kJ g–1
(ignore sign)
H2.
Energy from 1 mol Fe =
(iii)
(c)
Bonds broken OR Melting is endothermic
m(H2) = nM = 2mol  2 g mol–1 = 4 g
570 kJ
Energy per mol (H2) =
= 285 kJ mol-1
2 mol
n(H 2 ) in 1 g =
m
1g
=
= 0.5 mol
M 2 g mol-1
Energy per g (H 2 ) =
285 kJ mol-1
= 0.5 mol ´ 285 kJ mol-1
2 g mol-1
= 143 kJ g -1
m(CH4) = nM = 1mol  16 g mol–1 = 16 g
m
1g
n(CH 4 ) =
=
= 0.0625 mol
M 16 g mol-1
Energy per g (CH 4 ) =
890 kJ mol-1
= 0.0625 mol ´ 890 kJ mol-1
16 g mol-1
= 55.6 kJ (or kJ g -1 )
H2 provides the most energy per gram of fuel
•
ONE step correct (ignore sign).
NCEA Level 2 Chemistry (90310) 2011 — page 4 of 4
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
3A
3M
3E